From: Gareth McCaughan Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: 05 Nov 1998 16:22:54 +0000 Bruno Langlois wrote: > Let P a monic polynomial in Z[X]. > We suppose that : > - the degree of P is even, > - for all positive integer n, P(n) is a square in Z. > Can we say that P is a square in Z[X] ? Yes. Use the binomial theorem to write root(p(X)) = q(X) + O(1/X), so that p(X) = q(X)^2 + O(X^{n-1}), where deg p = 2n. For X very large such that p(X) isn't equal to q(X)^2, |p(X)-q(X)^2| >= q(X)^2 - [q(X)-1]^2 = 2q(X)-1. This is impossible for large X, since we just showed the LHS can't grow that fast. Hence, for all large enough X, p(X) = q(X)^2. Hence the polynomial p-q^2 has infinitely many zeros, and is therefore identically zero. [] -- Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics, gjm11@dpmms.cam.ac.uk Cambridge University, England. ============================================================================== From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: 5 Nov 1998 17:13:05 GMT In article <36419B77.1C1A@club-internet.fr>, langlois bruno wrote: >Let P a monic polynomial in Z[X]. >We suppose that : >- the degree of P is even, >- for all positive integer n, P(n) is a square in Z. >Can we say that P is a square in Z[X] ? Yes, it must be a square. If p(z) is monic with degree 2n, consider the Laurent series of q(z) = z^n sqrt(p(z)/z^(2n)) (taking the principal branch of the square root) on |z| > R = max{ |r|: p(r)=0 }. If p(z) is not a square, q(z) is not a polynomial, and we have q(z) = r(z) + a_k z^(-k) + O(|z|^(-k-1)) where r(z) is a polynomial with rational coefficients and a_k \ne 0. If M is the least common multiple of the denominators of the coefficients in r(z), M (q(z)- r(z)) can't be an integer when z is a sufficiently large integer, so neither can sqrt(p(z)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: Mark Sapir Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: Thu, 05 Nov 1998 12:01:25 -0600 langlois bruno wrote: > > Let P a monic polynomial in Z[X]. > We suppose that : > - the degree of P is even, > - for all positive integer n, P(n) is a square in Z. > Can we say that P is a square in Z[X] ? It can be a nice high school olympiad problem. Here is a sketch of an elementary solution: if f(x) has degree 2n, find a polynomial g(x) such that g(x)^2 has the same first n+1 coefficients as f(x) (it is always possible to do); g(x) will have ratioanal coefficients, which is not a problem, just multiply the variable x by a proper number, and you get a polynomial with integer coefficients. Then use the fact that the distance between two integer squares a^2-b^2 cannot be much smaller than 2*min{a,b}. -- Mark Sapir Professor of Mathematics Vanderbilt University Nashville, TN 37240 msapir@math.vanderbilt.edu (615)322-6657 ============================================================================== From: rusin@math.niu.edu (Dave Rusin) Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: 5 Nov 1998 17:40:54 GMT In article <36419B77.1C1A@club-internet.fr>, langlois bruno wrote: >Let P a monic polynomial in Z[X]. >We suppose that : >- the degree of P is even, >- for all positive integer n, P(n) is a square in Z. >Can we say that P is a square in Z[X] ? Yes. When I ran into this question last year I got this information from Bruce Reznick: >Michael Filaseta [...] contacted Schinzel, who'd know. The >consensus is that the original result (f: Z -> Z^2 => f = g^2) is >apparently well-known and has been discovered several times. I don't have any references, however. Reznick's solution was to compute Q = sqrt(P) = F + G formally where F is polynomial and G(n)->0 as n->oo. By assumption the values of Q are integers so the iterated differences are too; on the other hand high-enough differences of F also vanish. Thus the differences of G are both integral and small, hence zero, making G itself both polynomial and small, hence zero (as n->oo, and hence identically). So Q is a polynomial. Here's an alternative proof Neil Dummigan and I worked out: factor P = P1^2*P2 with P2 square-free; then y^2=P2(x) is a non-singular curve with infinitely many integral points, hence P2 is at most quadratic. (Obviously P2 can't be linear with all P2(n) squares unless P2=constant.) Completing the square makes this equivalent to an equation y^2=ax^2+b (Pell's equation) and the premise becomes that it has O(N) solutions with x Z, does it follow that P = f o g for some polynomial g? I suppose something like Reznick's solution will apply. Note that there is a distinction between integer polynomials and polynomials all of whose values at integers are integers (e.g. x(x+1)/2) so the question has to be phrased carefully. Note that the proofs cannot be purely formal, e.g. P(X) = X^4+4 is not a square in R[X] even though P(x) is a square in R for every x in R, when R = Z/5Z or R is the real number field. dave ============================================================================== From: Drew Vandeth Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: Fri, 06 Nov 1998 10:30:15 +1100 langlois bruno wrote: > Let P a monic polynomial in Z[X]. > We suppose that : > - the degree of P is even, > - for all positive integer n, P(n) is a square in Z. > Can we say that P is a square in Z[X] ? There is a classical theorem which states that if a polynomial with integral coefficients is an $m$th power for every integral value of its argument, then it is the $m$th power of a polynomial with integral coefficients. This follows from Hilbert's irreducibility theorem but can be proven in an elementary way, see for example P\'olya-Szeg\"o, Problems and theorems in analysis, vol II, part VIII, Chapter 2, Springer-Verlag 1976. -- Drew Vandeth ceNTRe for Number Theory Research Department of Mathematics Macquarie University Sydney, Australia vandeth@mpce.mq.edu.au ============================================================================== From: israel@math.ubc.ca (Robert Israel) Newsgroups: sci.math.research Subject: Re: Square of a polynomial ? Date: 10 Nov 1998 22:23:18 GMT In article <71tcso$od1$1@nnrp1.dejanews.com>, johnmitchell2100@my-dejanews.com writes: |> In article <36419B77.1C1A@club-internet.fr>, |> blang@club-internet.fr wrote: |> > Let P a monic polynomial in Z[X]. |> > We suppose that : |> > - the degree of P is even, |> > - for all positive integer n, P(n) is a square in Z. |> > Can we say that P is a square in Z[X] ? |> Several readers have already supplied proofs that this is true. I just wanted |> to mention that this question appeared on the "Aggregation" exam given to |> students at the Ecole Normale de Jeune Filles (in Paris) around 1980. One of |> the "jeunes filles" told an Argentinian mathematician I knew about the |> problem, and he then asked me about it. I came up with a finite difference |> solution that's essentially the same as Bruce Resnick's solution mentioned in |> Dave Rusin's reply. Notice that with this solution (and perhaps with others), |> you don't need the assumptions that P is monic and has even degree. On the other hand, some of these solutions (including mine, which used Laurent series) solve the harder question where "for all positive integer n" is replaced by "for infinitely many positive integers n". For that one you do need P to be monic and of even degree, otherwise counterexamples are P(n) = n and P(n) = 2 n^2 + 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2